What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?

Continuous functions have to be differentiable somewhere.
An infinitely differentiable function is analytic.
The Borel set of R is P(R).
There can't exist a continuous curve filling the entirely filling the square [0,1]x[0,1]

Linear imply continuous.

These 2 have my interests peaked.
Please do go on.
[citation needed]

>One can design a clever algorithm which can decide whether a general Diophantine equation is solvable or not

a.k.a Hilbert's 10th problem. Hint: such an algorithm is impossible.

Discontinuous linear functions require the axiom of choice AFAIK, so it's Banach-Tarski-tier shit.

>You can have a continuous function that takes a 3D vector, and produces a perpendicular vector to that.

You can't. See the hairy ball theorem.

discontinuous w.r.t. what topology? Surely this doesnt work in Euclidean space with standard topology?

In fact, differentiation is linearization of a function, so a linear functions shouldnt just be continuous, but even differentiable (everywhere)

that is not the statement of the hairy ball theorem.

>peaked
You fucking illiterate mongoloid, please stop posting.

It's a corollary.

peeked? piqued?

Please note that for many posters, English is not their first language (like Spanish guys, Germans, Americans etc.)

1)en.wikipedia.org/wiki/Cantor_function

2)f(x)=0 on ]-inf;0]
f(x)=exp(-1/x) on ]0; +inf]

3) en.wikipedia.org/wiki/Borel_set#Non-Borel_sets

gonna need some latex for the other 2, give me some time.

>en.wikipedia.org/wiki/Cantor_function
oops I mean fr.wikipedia.org/wiki/Fonction_de_Weierstrass
my bad.

but the HBT is about cont. vectorfield on spheres. For example it's false on doughnuts.

I mean, isn't the cross product trivially continuous if you just fix one argument? So given any vector v, just take the function to be cross product with v

Okay. Write that function.

f(v) = v x ??

What do you take the cross-product with? You can't have a constant vector, because that'll result in a zero vector when they're colinear.

no the function is f(w)=vxw which is continuous in w and perpendicular to v at every point

Let me clarify: we need a function that takes a nonzero vector v, and returns a nonzero vector that's perpendicular to v. It's impossible to have such a function that's continuous.

>You fucking illiterate mongoloid, please stop posting.
see

4) math.stackexchange.com/questions/141958/why-does-the-hilbert-curve-fill-the-whole-square

5)en.wikipedia.org/wiki/Discontinuous_linear_map

Also I remembered one who fucked with my mind at an exam when I was still a student:

The inverse function of a continous function is not always continuous.

No, even with usual topology, on infinite dimension space, linear doesn't imply continuous.

>The inverse function of a continous function is not always continuous.
Wouldn't trig functions and their inverses make that obvious

Trigonometric functions on R are not invertible. Restrict them to a principal branch and they become invertible, with continuous inverses.

The inverse function not being continuous thing may not be logically intuitive, but finding an example is extremely easy. x -> 1/x.

What? f(x) = 1/x is not continuous at x=0.
The theorem was: there's a continuous function whose inverse is discontinuous.

Except you only define the inverse trig functions in a certain interval to begin with, on which they are continuous?

Where did you get that pic op

I like it

It's only "obvious" to a first year that it will have simple behaviour.
Past first term even you already know that it's very easy to make nasty ODEs that do not have an easy to find solution.

>This, in general, is not true
No one claimed that the FTA worked in other domains.

>Continuous functions have to be differentiable somewhere.
Said no one ever. In general differentiable doesn't even imply continuous.

>The inverse function of a continous function is not always continuous.
You don't need to much to know that this isn't true... homeomorphisms require a continuous inverse for a reason.
If you map a line to a circle then you can easily see that it does not have a continuous inverse.

Wow your autism is showing, the thread is about intuitively true statement that are false. Stop pretending you knew this statments were false the first time you saw them especially when you say something as retarded as :
>In general differentiable doesn't even imply continuous.

no you need to go in higher dimension than 1 to find a counter exemple, that's why you may think it's true if you only think in dim 1.

t->exp(it) t in [0;2pi[
is the simplest example, the inverse is not continuous at 1. (because the neighborhood as images as low as 0 and as high as 2pi)

Why not? Say v=(v1,v2,v3). Set w=(-v2,v1,0). Done.

After your first term class in analysis you should already know to be very careful with things in a more generalised setting. If being autistic means that I'm not as retarded as you to keep making the same intuitive mistake, then sure, I'll take it.
>In general differentiable doesn't even imply continuous.
And this is retarded why? Do you really still believe that this is true?

Parallel postulate
Axiom of choice
Continuum hypothesis

For v=(0,0,1), this fails.

>Parallel postulate
whut

I agree completely with this.
Great mathematicians of the past have fallen for these subtly "obvious" truths that turned out to be false.
For example, there is a quick way to prove Fermats last theorem if we assume that integers of number fields generally have the fundamental theorem of arithmetic. It is generally believed that this was the flawed proof Fermat found.

Math is very subtle and it's just arrogant to say all of these statements are immediately obvious.

>parallel postulate
is obviously true

>axiom of choice
is an axiom, so neither true nor false.

>Continuum hypothesis
hm, not sure how the word "obvious" applies to reasonings about infinities that are larger than other infinities

>Great mathematicians of the past have fallen for these subtly "obvious" truths that turned out to be false.
This doesn't make you great or even close to those mathematicians.
>great mathematicians took a shit, used thihng toilet paper and smelt their hands afterwards before going to work on their latest work
>I do this too so I must be worthy of spitting out something great

>mad
You are an unpleasant person to communicate with, and I wish you would leave.

>What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?

The convergent series of continuous functions is continuous. Cauchy will never live that one down.

The intermediate value theorem. Seems to be true, but in fact requires an infinite amount of work to find the said point.

>theorem
>not true
ummm...

>mommy mommy he said something I didn't like

but that is obviously false if you know about the fourier series of the rectangle function, for example

Because being differentiable is a property way stronger than continuous. Differentiable imply continuous.

>The inverse function of a continous function is not always continuous.
yeah and this stems from the fact that inverse set functions do not commute with the conjunctions operator.

Did you read?
>In general
You realise that there are many different kinds of differentiability right? Not just the baby one variable one that you learn in babby's first real analysis class?
Learn to read properly.

>No one claimed that the FTA worked in other domains.
Not true at all. People tried to use this to prove Fermat's Last Theorem, thus prompting the work of Dedekind.

As I've said, it's impossible, but keep trying.

Hint: your solution fails for every (0,0,x) vector.

Hello Wildberger!

Back then yes, but nowadays I don't think so.

Do you even understand the thread? Are you an idiot?

>sphere
>nonmeasurable

>What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?
None of those things appear "obviously" true so don't even belong in the thread.

I don't think you get this thread topic. Just because it is currently taught to people who understand the theory doesn't make something not intuitive. Do you know why the people who tried proving FLT long ago thought FTA would apply to things like Z[zeta_3]? It's because it appears perfectly logical at first glance. Only by digging deeper do we see that these things are not inherently true,

There is only one definition of diffentiability, F is differentiable at x if :
F(x+h)=F(x)+L(h)+o(h)
With L linear and continuous.
It's obvious from the definition that F is continous at x. Dimensionality has nothing to do with the definition.

There are examples of functions that are Gateaux differentiable but not continuous, so yes, dimensionality matters.
For example, the function [math]f: mathbb{R}^2 \rightarrow \mathbb{R} [/math] defined by:
[eqn]f(x,y) = \begin{cases} \frac{x^2y}{x^4+y^2}, & \text{if } (x,y) \neq (0,0)\\ 0, & \text{if } (x,y) = (0,0) \end{cases} [/eqn]
is Gateaux differentiable at the origin but is discontinuous there.
To think there is only one definition of differentiability is naive. The condition that you stated is something that an ideal derivative would satisfy, but not all of them do.

>Gateaux differentiable
Do you have any idea why we call it Gateaux Differentiable and not differentiable?
Because it's not differentiable. Differentiable umply continuous, always.

Because it was named after René Gâteaux and for no other reason.
The "better" derivative, the Fréchet derivative, does satisfy the "differentiable implies continuous" that we want, but no one calls it just "derivative" because there are too many different kinds of derivatives that you could be talking about.

>is an axiom, so neither true nor false.

I can make an axiom that everything is made of earth, wind, fire, and water, or that the earth is 6000 years old.

Those axioms are obviously false.

You've misunderstood axioms then. You can't state an axiom and then appeal to reality to "disprove" it.

>I can make an axiom that everything is made of earth, wind, fire, and water, or that the earth is 6000 years old.
Yes you could, and those that believe to choose this axiom would build on it and consider results with respect to this axiom.
Those that disagree will pick another axiom.
But as has said, you've misunderstood what an axiom is.
This is similar to those that say "you can make theories so why can I make a theory about god and adam and eve?"

>retarded highschooler detected

Do you understand the thread? Nobody is sitting around saying "Oh! Gateau differentiable functions need not be continuous. I bet there is a function that is nowhere differentiable in the sense of single variable calculus!"

Nobody is sitting around saying that differentiable implies continuous in a general setting either.

you realize that in their respective fields, these are each taken as axioms, and are independent of the other "naive" axioms (C and CH are independent of ZF, PP is independent of other euclidean axioms)

You are literally stupid.
He knows they aren't true. They seem like they would be true.
As for the continuous implies differentiable somewhere: for a long time people thought this you fucking hick. It was revolutionary when it was shown to be untrue.

>They seem like they would be true.
What the actual fuck. Perhaps it would seem true to the uneducated regulars of Veeky Forums.

>The inverse function of a continuous function is not always continuous.

Can you give an example of an invertible [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math] with this property?

math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous

I saw that, and not one example is a function [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math].

Oh, I thought the first example would have been good enough for you. You might have a hard time finding something which goes from all of [math]\mathbb{R}[/math] to [math]\mathbb{R}[/math].

assuming arithmetic is consistent!

I want an example with my stated property because, when the topologies can be different, it's actually quite obvious (e.g. biject [0,1) with a circle in the obvious way).

But when both domain and range are the standard [math]\mathbb{R}[/math], that is when it goes against intuition.

It is clear that any example of a continuous bijection [math] f: \mathbb{R} \rightarrow \mathbb{R} [/math]that is not bicontinuous cannot be monotone, so it must look pretty funny. I'll see if I can either think of an example or prove there is none.

As , I realize it's obvious that there is none.

This follows from my comment together with proofwiki.org/wiki/Continuous_Function_on_Closed_Interval_is_Bijective_iff_Strictly_Monotone

So the inverse of any continuous bijection [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math] is also continuous.

Sorry, I see additional conditions were added after . As originally stated, even the zero map would work.

I agree this is a good example for the thread.

The sum of all natural numbers is -1/12

how is that obvious?

>but finding an example is extremely easy. x -> 1/x.

1/x is continous everywhere. but its not the inverse function of x -> x, because x->x is defined for x = 0 as well

>because x->x is defined for x = 0 as well
The inverse of x -> x is itself..

>Americans

hearty out loud lol

That the multiplicative group of the complex unit circle is not isomorphic to the multiplicative group [math]\mathbb{C}^\ast[/math].

The two actually are isomorphic.

differentiation operators are linear and not continuous (you can't just swap limits with differentiation most of the time)
they don't require axiom of choice
linear + bounded implies continuous, which is a given for finite dimensional linear operators

Here are some things that are "obviously true", but which can fail in the absence of the axiom of choice:

1) Any two algebraic closures of [math]\mathbb{Q}[/math] are isomorphic.

2) Every field has an algebraic closure.

3) Every vector space has a basis.

4) If a real number is in the closure of a set [math]X \subset \mathbb{R} [/math], then it is the limit of a sequence from [math]X [/math].

5) The reals are not a countable union of countable sets.

6) If a set is infinite, it has a countably infinite subset.

7) If you have an equivalence relation on [math]\mathbb{R}[/math], then the number of equivalence classes is no larger than the cardinality of [math]\mathbb{R} [/math].

do you have proofs of 1, 4, 6 and 7 failing? 5 is always true: the reals are never countable

1^3+0^3=1^3
0^5+0^5=0^5

Hah! I was hoping someone would assert that 5 is always true.

No, 5 is not always true. The reason is that, in the absence of AC, a countable union of countable sets need [math]not[/math] be countable. The reals actually can be a countable union of countable sets. Therefore, in particular, the theory of Lebesgue measure can fail totally.

Proofs of the consistencies of the negations of what you asked require sophisticated forcing and inner model theory arguments. Refer to Jech's two textbooks "Set Theory" and "The Axiom of Choice", among others. If you'll accept an argument from authority, see mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice/70435#70435.

Here's one that I sort of implicitly assumed for a couple years in my visualization until I realized it to be false, even in the most trivial case:

"If [math]N[/math] is a transitive model of ZFC, [math]\kappa = \text{ORD}^N[/math], [math]M[/math], and [math]M \supset N [/math] is minimal such that [math]\kappa \in M[/math], then [math]\kappa[/math] is inaccessible in [math]M[/math]."

More informally, if one has a model [math]N[/math] whose class of ordinals is [math]\kappa[/math], then if one "keeps climbing" up the ordinal ladder past [math]\kappa[/math], adding precisely those necessary sets mandated by ZFC along the way, then when one "catches up" to a model of ZFC, [math]\kappa[/math] will be inaccessible.

_____

In fact, even if one deals strictly with models of V=L, it can be the case that [math]\kappa[/math] is forced to be not only not inaccessible, but countable. So sets one must constructively add higher up can collapse the entire model.

I happened to type up a proof here .

no surprise at all.

In fact, the following seems intuitively true
>Given non-empty sets, their cartesian product will again be non-empty

but guess what..

>If a set is infinite, it has a countably infinite subset.
Can't we go further ? and say you can't be sure it has non trivial subsets (null and himself) without AC.

Well, the assertion that the cartesian product of nonempty sets is nonempty is clearly equivalent to AC; lists predominantly very non-trivial possibilities in the absence of AC.

No.

However, of note is that without AC there can exist an infinite Dedekind-finite sets. A set [math]X[/math] is "Dedekind finite" if there is no proper injection [math]f: X \rightarrow X [/math]. Intuitively, and in the presence of AC, the Dedekind-finite sets are precisely the finite sets. However, in the absence of AC, it is possible for there to exist infinite Dedekind-finite sets.